(i) The width of depletion layer decreases.
(ii) The width of depletion layer increases

A rectifier is used to convert alternating current into direct current, whose labelled circuit is given below.

Working
During the positive half cycle of the input AC, the diode D₁, is forward biased and the diode D₂ is reverse biased. The forward current flows through diode D₁. 
During the negative half cycle of the input AC, the diode D₁ is reverse biased and diode D₂ is forward biased. Thus, current flows through diode D₂ . Thus, we find that during both the halves, current flows in the same direction.

Voltage gain \(=-trans-conductance\times output\ resistance \)

In junction diode, the direction of diffusion current is from P-region to N-region.

The fermi-level of intrinsic semiconductor lies mid way between its valence and conduction bands i.e., in the middle of forbidden energy gap of intrinsic semi-conductor.

Two processes that takes place during the formation of p-n junction are diffusion and drift of charge carriers. In an n-type semiconductor, the concentration of electrons is more than that of holes. Similarly, in a p-type semiconductor, the concentration of holes is more than that of electrons. Formation of depletion region during formation of p-n junction and due to the concentration gradient across p and n-sides, holes diffuse from p-side to n-side
(p ⟶ n) and electrons diffuse from n-side to p-side (n ⟶ p). The diffused charge carriers combine with their counterparts in the immediate vicinity of the junction and neutralise.

This sets up potential difference across the junction and an internal electric field E; directed from n-side to p-side. The equilibrium is established when the field E; becomes strong enough to stop further diffusion of the majority charge carriers (however, it helps the minority charge carriers to drift across the junction). The region on either side of the junction which becomes depleted (free) from the mobile charge carriers is called depletion region or depletion layer. The potential difference developed across the depletion region is called the potential barrier.

Rectification efficiency
\(=\frac { d.c.output \ power }{ a.c.input \ power } \times100\)
\( =\frac { 40 }{ 100 } \times100=40%\)
Power efficiency 
 \(=\frac { d.c.output \ power }{ a.c.input \ power \ for \ half \ cycle } \times100\) 
\( =\frac { 40 }{ (100/2) } \times100=80%\)

Metallic solids are used to produce good conductors. Covalent solids are generally used to produce semiconductors and the ionic and van der Waal's solids are used to produce insulators.

\({ \left( 9.25 \right) }_{ 10 } \ = \ { \left( 9 \right) }_{ 10 }+{ \left( 0.25 \right) }_{ 10 }\)       
 \(=\ { \left( 1001 \right) }_{ 2 }+{ \left( 0.01 \right) }_{ 2 }\)
 \(=\ { \left( 1001.01 \right) }_{ 2 }\)

\(\beta_{a c}=\left(\frac{\Delta I_{C}}{\Delta I_{B}}\right)_{V_{C E}}, \quad \beta_{d c}=\frac{I_{C}}{I_{B}}\)
For determining \(\beta\)_ac and \(\beta\)_dc at the stated values of V_CE and I_C one can proceed as follows. Consider any two characteristics for two values of I _B which lie above and below the given value of I_C. Here I_C = 4.0 mA. (Choose characteristics for I_B= 30 and 20 \(\mu\)A.) At V_CE = 10 V we read the two values of I_C from the graph. Then
\(\Delta I_{B}=(30-20) \mu \mathrm{A}=10 \mu \mathrm{A}, \Delta I_{C}=(4.5-3.0) \mathrm{mA}=1.5 \mathrm{~mA}\)
Therefore, \(\beta_{a c}=1.5 \mathrm{~mA} / 10 \mu \mathrm{A}=150\)  
For determining \(\beta\)dc, either estimate the value of I _B corresponding to I _C = 4.0 mA at V_CE = 10 V or calculate the two values of \(\beta\)dc for the two characteristics chosen and find their mean.
Therefore, for I _C = 4.5 mA and I_B = 30 \(\mu\)A
\(\beta_{d c}=4.5 \mathrm{~mA} / 30 \mu \mathrm{A}=150\)
and for I_C = 3.0 mA and I_B = 20 \(\mu\)A
\(\beta_{d c}=3.0 \mathrm{~mA} / 20 \mu \mathrm{A}=150\)
Hence, \(\beta_{d c}=(150+150) / 2=150\)

(a) The circuit diagram is shown here:

The emitter-base junction, being forward biased, the majority charge carriers (electrons), from the emitter, flow into the base region constituting the emitter current (I_E) The base region, being very thin, only a (very) small fraction, of these (small base current (I_B). 1 The majority of these charge carriers, are attracted by the (reverse biased) collector. These make up the collector current (I_C )
I_E = I_C + I_B
The circuit diagram of a transistor, working as an amplifier, in its C _E mode, is shown here. 

If a small sinusoidal voltage is superimposed on the dc base bias by connecting the source of this signal in series with V_BB supply. Then the base current will have sinusoidal variations superposed on the values I_B. as a consequence the collector current also will have sinusoidal variation superimposed on the value of I_C producing in turn corresponding change in the output voltage V₀ .

(a) Due to the diffusion of electrons and the holes, from their majority zone to minority zone, a layer of positive and negative space charge region on either side on the junction is formed. This is called the depletion region.
The loss of electrons, from n-region and gain of electrons by the p-region, causes a difference of potential across the junction. This tends to prevent the movement of charge carriers across the junction and is, therefore, termed as barrier potential.

For positive half cycle of input ac, one of the two diodes gets forward biased and conducts and output current is obtained across the load RL, For negative half cycle of input ac, the other diode
gets forward biased and thus output current is obtained due to it. Therefore, output is obtained for both the cycles of input ac.

(a) Emitter: It is of moderate size and heavily doped 
Base: It is very thin and lightly doped Collector : It is moderately doped and larger in size.
(b) Transistor is said to be in active state when its emitter-base junction is (suitably) forward biased and base-collector junction is (suitably) reverse biased.

(c)When a small sinusoidal voltage is superposed on the dc base biased, the base current will have sinusoidal variation superimposed on the value of I_E. As a consequence, the collector current also will have sinusoidal variations, superimposed on the
Value of I_C, producing corresponding (amplified) changes in the value of V_{0 .} 
ac current gain, \({ \beta }_{ ac }={ \left( \frac { \Delta { I }_{ C } }{ \Delta { I }_{ B } } \right) }_{ { V }_{ CB } }\)

(i) We cannot measure the potential barrier of \(p-n\) junction by putting a sensitive voltmeter across its terminals because there are no free electrons or holes in the depletion layer. And in the absence of forward biasing, the depletion layer offers infinite resistance.
(ii) From the above study,we find that when some move supports the tendency of majority of the people, the resistance/(opposition) is lowered and output current (desired result) is large (appreciable) However, when some move is against the tendency of majority of people, it suffers rough weather.

(i) (b) : Voltage drop across R.
\(V_{R}=V_{B}-V_{N}=5-0.7=4.3 \mathrm{~V}\)
Here, I_min = 1 x 10^-3 A
\(R_{\max }=\frac{V_{R}}{I_{\min }}=\frac{4.3}{1 \times 10^{-3}}=4.3 \times 10^{3} \Omega=4.3 \mathrm{k} \Omega\)
(ii) (b) : \(I=6 \mathrm{~mA}=6 \times 10^{-3} \mathrm{~A}\)
\(\begin{array}{l} V_{R}=V_{B}-V_{N}=5-0.7=4.3 \mathrm{~V} \\ R=\frac{V_{R}}{I}=\frac{4.3}{6 \times 10^{-3}}=717 \Omega \end{array}\)
(iii)(d) : Here, V_B = 6 V; V_N = 0.7V,
\(V_{R}=6-0.7=5.3 \mathrm{~V}\)
Power dissipated in R = l x V_R
= (6 x 10^-3) x 5.3 = 31.8 x 10^-3 W
= 31.8 mW
(iv) (b) : V_t = V_bi + V_R = 0.63 + 6 = 6.63 V
(v) (d) : Diode is two terminal device, anode and cathode are the two terminals.